Question:
In a triangle \(ABC\), if \(r_1>r_2>r_3\), then which of the following is true?
In a triangle \(ABC\), if \(r_1>r_2>r_3\), then which of the following is true?
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The exradius opposite side \(a\) is \(r_1=\frac{\Delta}{s-a}\). Larger exradius means a smaller denominator \(s-a\), which in turn means a larger corresponding side.
Updated On: Jun 18, 2026
- \(a>b>c\)
- \(a>b,\; b<c\)
- \(a<b<c\)
- \(ac\)
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The Correct Option is A
Solution and Explanation
Step 1: Recall the exradius formulas.
For a triangle with area \(\Delta\) and semi-perimeter \(s\), \[ r_1=\frac{\Delta}{s-a}, \] \[ r_2=\frac{\Delta}{s-b}, \] \[ r_3=\frac{\Delta}{s-c}. \]
Step 2: Use the given inequality.
Given, \[ r_1>r_2>r_3. \] Substituting the formulas, \[ \frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}. \] Since \(\Delta>0\), we obtain \[ \frac{1}{s-a} > \frac{1}{s-b} > \frac{1}{s-c}. \] Therefore, \[ s-a<s-b<s-c. \]
Step 3: Compare the sides.
Subtracting \(s\) from each part, \[ -a<-b<-c. \] Multiplying by \(-1\), \[ a>b>c. \]
Step 4: Final conclusion.
Hence, \[ \boxed{a>b>c} \]
For a triangle with area \(\Delta\) and semi-perimeter \(s\), \[ r_1=\frac{\Delta}{s-a}, \] \[ r_2=\frac{\Delta}{s-b}, \] \[ r_3=\frac{\Delta}{s-c}. \]
Step 2: Use the given inequality.
Given, \[ r_1>r_2>r_3. \] Substituting the formulas, \[ \frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}. \] Since \(\Delta>0\), we obtain \[ \frac{1}{s-a} > \frac{1}{s-b} > \frac{1}{s-c}. \] Therefore, \[ s-a<s-b<s-c. \]
Step 3: Compare the sides.
Subtracting \(s\) from each part, \[ -a<-b<-c. \] Multiplying by \(-1\), \[ a>b>c. \]
Step 4: Final conclusion.
Hence, \[ \boxed{a>b>c} \]
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